为什么哈希函数对hascode执行XOR？

Question

我阅读了说明，但是我无法理解通过对hashCode进行XOR所实现的目标。谁能举个例子。

static final int hash(Object key) {
    int h;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

此代码摘自HashMap源代码。我只是想知道为什么他们使用XOR，Marko正确地回答说HashMap实现使用低端位。我认为不仅HashMap，其他集合也将这样做，这就是为什么我没有提到任何集合名称的原因。我不明白为什么人们会“贬低”这个问题。

Answer 1

这是一种防止“错误”哈希码的典型方法：此类低端位的变量不够可变。 Java的HashMap实现仅依赖哈希码的低端位来选择存储桶。

但是，此代码的动机已经很久了，因为HashMap已经进行了自己的扩展。如果在Hashtable上使用它会很有意义，但是当然，自2000年以来编写的任何代码都不应使用它。

Answer 2

该代码是openjdk java HashMap源代码：HashMap.java

     /**
     * Computes key.hashCode() and spreads (XORs) higher bits of hash
     * to lower.  Because the table uses power-of-two masking, sets of
     * hashes that vary only in bits above the current mask will
     * always collide. (Among known examples are sets of Float keys
     * holding consecutive whole numbers in small tables.)  So we
     * apply a transform that spreads the impact of higher bits
     * downward. There is a tradeoff between speed, utility, and
     * quality of bit-spreading. Because many common sets of hashes
     * are already reasonably distributed (so don't benefit from
     * spreading), and because we use trees to handle large sets of
     * collisions in bins, we just XOR some shifted bits in the
     * cheapest possible way to reduce systematic lossage, as well as
     * to incorporate impact of the highest bits that would otherwise
     * never be used in index calculations because of table bounds.
     */
    static final int hash(Object key) {
        int h;
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }

xor是为了让散列结果更加分散